Question

Three rods A, B and C form an equilateral triangle at $0^{o}C$. Rods AB and BC have same coefficient of expansion ${\alpha }_1$ and rod AC has ${\alpha }_2$. If change in angle between AB and BS is $d\theta$ than temperature through which system is heated.

• A. $\frac{\sqrt{2d\theta }}{3({\alpha }_2-{\alpha }_1)}$
• B. $\frac{\sqrt{3d\theta }}{2({\alpha }_2-{\alpha }_1)}$
• C. $\frac{\sqrt{2d\theta }}{5(2{\alpha }_2-{\alpha }_1)}$
• D. $\frac{\sqrt{3d\theta }}{({\alpha }_2-{\alpha }_1)}$

Answer 1

The correct option is B $\frac{\sqrt{3d\theta }}{2({\alpha }_2-{\alpha }_1)}$

REF.Image.

Let $AB=I_1$

$BC=I_2$

$CA=I_3$

using cosine's formula,

$\therefore cos\theta =\frac{I_2^2+I_1^2-I_3^2}{2I_2{I}_1}$

$2I_2{I}_{1}cos\theta =I_2^2+I_1^2-I_3^2...\left(i\right)$

on differentiating the equation (i) we get

$2(I_{2}d{I}_1+I_{1}d{I}_2)cos\theta -2I_1{I}_{2}sin\theta d\theta =2I_{2}d{I}_1+2I_{1}d{I}_1-2I_{3}d{I}_1$

Here, $(I_1=I_2=I_3=I)$ for equilateral triangle

$d{I}_1=I{\alpha }_1△t,d{I}_2=I{\alpha }_1△t$

$d{I}_3=I{\alpha }_2△t$

$\therefore 2(I^2{\alpha }_1△t+I^2{\alpha }_1△t)cos\theta -2I^{2}sin\theta d\theta$

$=2I\times I{\alpha }_1△t+2I\times I{\alpha }_1△t-2I\times I{\alpha }_2△t$

Dividing by $2I^2$we get,

$4{\alpha }_1△tcos\theta -2sin\theta d\theta =4{\alpha }_1△t-2{\alpha }_2△t$

$sin\theta d\theta =2{\alpha }_1△t(cos\theta -1)+{\alpha }_2△t$

putting $\theta ={60}^{\circ }$ (for equilateral triangle)

$d\theta \times sin{60}^{\circ }=2{\alpha }_1△t(cos{60}^{\circ }-1)+{\alpha }_2△t$

$d\theta \times \frac{\sqrt{3}}{2}=({\alpha }_2△t-{\alpha }_1△t)$

$△t=\frac{\sqrt{3}d\theta }{2({\alpha }_2-{\alpha }_1)}$